Trigonometric Arithmetic V

Rewrite $\sum_{k=0}^{n-1}\cos\left(e+k\frac{2\pi}n\right)$ which is the real part of $\operatorname{Re}\sum_{k=0}^{n-1}e^{i\left(e+2k\pi/n\right)}=\operatorname{Re} \ \ e^{ie} \sum_{k=0}^{n-1}e^{i2k\pi/n}$ And note that the last sum is a geometric sum. Letting $z=e^{2\pi i /n}$ we have $\sum_{k=0}^{n-1}z^k=\frac{1-z^n}{1-z}$ and note how this cancels to $0$ as $e^{i2\pi}=1$ . Thus the final answer is $\boxed{0}$ .

That's really cute, this is the same as adding all the complex roots of 1, rotated by an arbitrary angle, which is e in this case. So, it all adds up to nothing. I had to sit and stare at this for a while before any of it made any sense.